# Number density of phonons in superfluid

I'm reading Superfluidity and Superconductivity by Tilley & Tilley. In section 2.4, the argument is made that the normal component of superfluid helium consists of phonons and rotons. The thermodynamic properties can be split into contributions from the two, and these then stem from a calculation of the respective number densities. These are given in equation (2.34), with $$N_{ph}$$ given as

$$N_{ph} = 9.60 \pi (\frac{k_B T}{hc})^3$$.

I am unsure where this comes from. It is argued that Bose distribution function is appropriate: does this calculation then boil down to

$$N_{ph} = \int \frac{1}{e^{\epsilon /T} - 1} d\epsilon$$

with the appropriate dispersion for phonons (i.e. $$\epsilon = cp$$ with c the speed of sound)? If so, I do not see how this reduces the given solution. Any help is appreciated.

Best, Jack.

The number of states of one particle in a tree-dimensional volume $$V$$ with momentum in $$d^3\vec{p}$$ is known to be equal to $$\frac{Vd^3\vec{p}}{h^3}$$ At temperature $$T$$, the mean number of bosonic quasiparticles in a state with momentum $$\vec{p}$$ and energy $$\varepsilon(\vec{p})$$ is given by the Bose-Einstein distribution $$\overline{N}_p = \frac1{e^{\varepsilon(\vec{p})/k_BT}-1}$$ Hence the concentration of phonons is $$n_{ph} = \frac1{h^3}\int\frac1{e^{\varepsilon(\vec{p})/k_BT}-1}d^3\vec{p},$$ where for phonons $$\varepsilon(\vec{p}) = cp$$, $$p\equiv|\vec{p}|$$. Transformation of the integral due to the spherical symmetry gives $$n_{ph} = \frac{4\pi}{h^3}\int\limits_0^\infty\frac{p^2}{e^{cp/k_BT}-1}dp.$$ Change of variables $$p = \xi\, k_BT/c$$ in the integral leads to the expression $$n_{ph} = \left(\frac{k_BT}{hc}\right)^3 4\pi\int\limits_0^\infty\frac{\xi^2}{e^\xi-1} d\xi.$$ So the number $$9.60$$ in the initial formula is obtained as $$4\int\limits_0^\infty\frac{\xi^2}{e^\xi-1} d\xi = 4\,\Gamma(3)\zeta(3) \approx 9.6.$$
• In derivations of thermodynamic quantities I often see the measure accompanied by a factor $(2\pi)^3$ , i.e. $\frac{d^3 \boldsymbol{p}}{(2\pi)^3}$, why is that not the case here? Commented Jan 28, 2020 at 9:26
• @JackHughes There are two Planck constants $\hbar$ and $h$, $h = 2\pi\hbar$. So $d^3\vec{p}/h^3 = d^3\vec{p}/(2\pi\hbar)^3$. In units were $\hbar = 1$ this formula becomes $d^3\vec{p}/(2\pi)^3$.
• should the final result for $n_{ph}$ not depend on volume as pointed out in your first line? Commented Feb 5, 2020 at 11:50
• @JackHughes, $n_{ph} = N/V$ is a concentration, i.e. a number of phonons in the unit volume. This quantity depends only on the temperature.